## LynFran one year ago tan(1/2(cos^−1(x))=√1−x/1+x

1. anonymous

Do you have any idea on starting this question?

2. LynFran

no is say prove that their equal

3. anonymous

Well that's the first step :) Do you know what to do next?

4. LynFran

no

5. anonymous

We don't need to rush in and do everything in one go :) We can firstly start of with: tan(1/2(cos^−1(x))

6. anonymous

Tan ( 1 ) 2 Cos^-1(x)

7. anonymous

Try changing the inverse trig ratio

8. anonymous

You get something like: Tan (Secx) 2

9. LynFran

ok

10. anonymous

So graphically, it looks something like this

11. anonymous

|dw:1433678743148:dw|

12. anonymous

The only way to get the prove this, is by using properties of Pythagoras's, since we are dealing with a right angle triangle

13. LynFran

ok so we finding the hypotenuse in other word?

14. anonymous

Yeah we could try to find the hypotenuse or the other side. Just some way to get us that Square root, to help us prove that LHS = RHS

15. LynFran

so we would have x^2=(1/cosx)^2+2^2)?

16. Loser66

Question: $$tan (\dfrac{1}{2} arccos (x))$$ or $$tan (\dfrac{1}{2arccos (x))}$$. which one?

17. LynFran

tan(12arccos(x))

18. LynFran

the first 1

19. Loser66

if so, then the right hand side must be $$\dfrac{\sqrt {1-x^2}}{1+x}$$ , right?

20. Loser66

you missed the square at x from numerator. Am I right? please check the problem

21. Loser66

|dw:1433681866420:dw|

22. LynFran

the right side has no square at x its just; the root of (1-x)/(1+x)

23. Loser66

If it is so, I failed. :)